When Ted walked down the down-moving escalator, he reached the bottom after taking 50 steps. As an experiment, he then ran up the same escalator, one step at a time, reaching the top after taking 125 steps. Assuming that Ted went up five times as fast as he went down (that is, took five steps to every one step before), and that he made each trip at a constant speed, how many steps would be visible if the escalator stopped running?

Let n be the number of steps visible when the escalator is not moving, and let a unit of time be the time it takes Ted to walk down one step. If he walks down the down-moving escalator in 50 steps, then n - 50 steps have gone out of sight in 50 units of time. It takes him 125 steps to run up the same escalator, taking five steps to every one step before. In this trip, 125 - n steps have gone out of sight in 125/5, or 25 units of time. Since the escalator runs at a constant speed, we have the following linear equation that readily yields a value for n of 100 steps: